A) \[y=\pm 3\]
B) \[x=\pm \sqrt{5}\]
C) \[y=0,\ y=6\]
D) None of these
Correct Answer: C
Solution :
Change the equation \[9{{x}^{2}}+5{{y}^{2}}-30y=0\] in standard form \[9{{x}^{2}}+5({{y}^{2}}-6y)=0\] Þ \[9{{x}^{2}}+5({{y}^{2}}-6y+9)=45\]Þ\[\frac{{{x}^{2}}}{5}+\frac{{{(y-3)}^{2}}}{9}=1\] \[\because {{a}^{2}}<{{b}^{2}},\] so axis of ellipse on y-axis. At y axis, put\[x=0\], so we can obtained vertex. Then \[0+5{{y}^{2}}-30y=0\]Þ\[y=0,\,\,y=6\] Therefore, tangents of vertex\[y=0,\,\,\,y=6\].
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